| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Tank/reservoir mixing problems |
| Difficulty | Standard +0.3 This is a straightforward integrating factor question with standard steps: rearrange to standard form, find integrating factor 1/(1+t), integrate to find general solution, apply initial condition, then evaluate at specific times. The algebra is routine and the context is typical. Slightly easier than average (0.0) as it follows a mechanical procedure with no conceptual surprises, though the square root term requires careful integration. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1+t)\frac{dP}{dt}+P = t^{\frac{1}{2}}(1+t) \Rightarrow \frac{dP}{dt}+\frac{P}{1+t}=t^{\frac{1}{2}}\) | B1 | Correct rearrangement; may be implied by subsequent work |
| \(I = e^{\int\frac{1}{1+t}dt} = 1+t \Rightarrow P(1+t) = \int t^{\frac{1}{2}}(1+t)\,dt = \ldots\) | M1 | Uses model to find integrating factor and attempts solution |
| \(P(1+t) = \frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+c\) | A1 | Correct solution |
| \(t=0,\ P=5 \Rightarrow c=5\) | M1 | Interprets initial conditions to find constant; must use \(t=0\) and \(P=5\) |
| \(P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} = \frac{\frac{2}{3}(8^{\frac{3}{2}})+\frac{2}{5}(8^{\frac{5}{2}})+5}{9} = \ldots\) | M1 | Uses solution to find population after 8 hours |
| \(= 10\,277\) bacteria (allow awrt 10 300) | A1 | cso; correct number from correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} \Rightarrow \frac{dP}{dt} = \frac{(1+t)(t^{\frac{1}{2}}+t^{\frac{3}{2}})-(\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5)}{(1+t)^2}\) | M1 | Differentiates model using appropriate method; trivialised differentiation from incorrect work is M0 |
| Correct differentiation of correct answer to (a) up to constant of integration | A1ft | Follow through on their \(c\); if implicit differentiation used must get to function in \(t\) only |
| \(\left(\frac{dP}{dt}\right)_{t=1} = \frac{5\times10-\left(\frac{16}{3}+\frac{64}{5}+5\right)}{(5)^2} = \frac{403}{375}\) | dM1 | Uses \(t=4\) in their \(\frac{dP}{dt}\) to obtain value |
| \(\frac{403}{375}\times 1000 = \frac{3224}{3}\) (= awrt 1070) bacteria per hour | A1 | Correct answer; allow 1075 or answers rounding down to 1070 with correct units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} = \frac{\frac{16}{3}+\frac{64}{5}+5}{(1+4)}\) | M1 | Substitutes \(t=4\) into their \(P\) |
| \(= \frac{347}{75}\) | A1ft | Correct value for \(P\); follow through on their constant from (a) |
| \((1+t)\frac{dP}{dt}+P = t^{\frac{1}{2}}(1+t) \Rightarrow 5\frac{dP}{dt}+\frac{347}{75} = 2\times5 \Rightarrow \frac{dP}{dt}=\frac{403}{375}\) | dM1 | Uses \(t=4\) and their \(P\) to find \(\frac{dP}{dt}\) |
| \(\frac{403}{375}\times1000 = \frac{3224}{3}\) (= 1075) bacteria per hour | A1 | Correct answer with correct units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. The number of bacteria increases indefinitely which is not realistic | B1 | Suggests suitable limitation referring to the model; allow sensible comment even without equation. Do not allow answers referring to external factors such as temperature |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+t)\frac{dP}{dt}+P = t^{\frac{1}{2}}(1+t) \Rightarrow \frac{dP}{dt}+\frac{P}{1+t}=t^{\frac{1}{2}}$ | B1 | Correct rearrangement; may be implied by subsequent work |
| $I = e^{\int\frac{1}{1+t}dt} = 1+t \Rightarrow P(1+t) = \int t^{\frac{1}{2}}(1+t)\,dt = \ldots$ | M1 | Uses model to find integrating factor and attempts solution |
| $P(1+t) = \frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+c$ | A1 | Correct solution |
| $t=0,\ P=5 \Rightarrow c=5$ | M1 | Interprets initial conditions to find constant; must use $t=0$ and $P=5$ |
| $P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} = \frac{\frac{2}{3}(8^{\frac{3}{2}})+\frac{2}{5}(8^{\frac{5}{2}})+5}{9} = \ldots$ | M1 | Uses solution to find population after 8 hours |
| $= 10\,277$ bacteria (allow awrt 10 300) | A1 | cso; correct number from correct equation |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} \Rightarrow \frac{dP}{dt} = \frac{(1+t)(t^{\frac{1}{2}}+t^{\frac{3}{2}})-(\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5)}{(1+t)^2}$ | M1 | Differentiates model using appropriate method; trivialised differentiation from incorrect work is M0 |
| Correct differentiation of correct answer to (a) up to constant of integration | A1ft | Follow through on their $c$; if implicit differentiation used must get to function in $t$ only |
| $\left(\frac{dP}{dt}\right)_{t=1} = \frac{5\times10-\left(\frac{16}{3}+\frac{64}{5}+5\right)}{(5)^2} = \frac{403}{375}$ | dM1 | Uses $t=4$ in their $\frac{dP}{dt}$ to obtain value |
| $\frac{403}{375}\times 1000 = \frac{3224}{3}$ (= awrt 1070) bacteria per hour | A1 | Correct answer; allow 1075 or answers rounding down to 1070 with correct units |
**Alternative for (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} = \frac{\frac{16}{3}+\frac{64}{5}+5}{(1+4)}$ | M1 | Substitutes $t=4$ into their $P$ |
| $= \frac{347}{75}$ | A1ft | Correct value for $P$; follow through on their constant from (a) |
| $(1+t)\frac{dP}{dt}+P = t^{\frac{1}{2}}(1+t) \Rightarrow 5\frac{dP}{dt}+\frac{347}{75} = 2\times5 \Rightarrow \frac{dP}{dt}=\frac{403}{375}$ | dM1 | Uses $t=4$ and their $P$ to find $\frac{dP}{dt}$ |
| $\frac{403}{375}\times1000 = \frac{3224}{3}$ (= 1075) bacteria per hour | A1 | Correct answer with correct units |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. The number of bacteria increases indefinitely which is not realistic | B1 | Suggests suitable limitation referring to the model; allow sensible comment even without equation. Do not allow answers referring to external factors such as temperature |\begin{enumerate}
\item A sample of bacteria in a sealed container is being studied.
\end{enumerate}
The number of bacteria, $P$, in thousands, is modelled by the differential equation
$$( 1 + t ) \frac { \mathrm { d } P } { \mathrm {~d} t } + P = t ^ { \frac { 1 } { 2 } } ( 1 + t )$$
where $t$ is the time in hours after the start of the study.\\
Initially, there are exactly 5000 bacteria in the container.\\
(a) Determine, according to the model, the number of bacteria in the container 8 hours after the start of the study.\\
(b) Find, according to the model, the rate of change of the number of bacteria in the container 4 hours after the start of the study.\\
(c) State a limitation of the model.
\hfill \mbox{\textit{Edexcel CP1 2020 Q7 [11]}}